Suppose that we have $\{X_{\alpha}\}_{\alpha \in J}$, an indexed family of topological spaces. Let $X := \prod_{\alpha \in J}X_{\alpha}$. When we have a map $f_{\alpha} : A \rightarrow X_{\alpha}$ with a topological space $A$. Define $f : A \rightarrow \prod_{\alpha \in J}X_{\alpha}$ by $a \mapsto (f_{\alpha}(a))_{\alpha \in J}$.

Property. We know that $f$ is continuous if and only if each $f_{\alpha}$ is continuous, once we are given product topology on $X$. We further know that this property does not generally hold with box topology on $X$.

After reviewing the definition, it seems to me that this property seems rather trivial (or natural) since product topology only collects finite intersections of inverse images under projections as open sets.

Can someone give more examples outside this property that explains why we prefer product topology? A good list of evident examples will be much preferred than statement of theorems.

This really is the important property. It says that the product topology is the categorical product in topological spaces (en.wikipedia.org/wiki/Product_(category_theory)). A general feature of categorical products, like all universal constructions, is that they are unique if they exist; consequently this property alone completely determines everything about the product topology.
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Qiaochu YuanOct 24 '12 at 21:45

Thanks for the comment. I don't know why but your link does not acknowledge one parenthesis in the end. It really makes sense that this is "THE" property that we want as continuous maps are the important maps that we want to think about given topological spaces. I still want some more naive examples that can stimulated me even more though.
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user123454321Oct 24 '12 at 21:55

3 Answers
3

I think compactness is such an important property that it is also worth mentioning here.

In product topology we are able to show that if $X_{i}$ is compact for all $i\in I$, then $\Pi_{i\in I}X_{i}$ is compact, no matter what the index set $I$ is. This is also known as the Tychonoff theorem.

Does this also hold for the box topology? Consider for example $X_{i}=\{0,1\}$ for all $i\in\mathbb{N}$ with the discrete topology. These are obviously compact since the topologies are finite. However, the set $\Pi_{i\in\mathbb{N}}X_{i}=\{0,1\}^{\mathbb{N}}$ is uncountable, and with box topology it has the discrete topology. Why? Because for any $x\in\{0,1\}^{\mathbb{N}}$ we have $\{x\}=\{(x_{1},x_{2},...)\}=\Pi_{i\in\mathbb{N}}\{x_{i}\}$, and this is an open set in the box topology, as the singletons $\{x_{i}\}$ are open in $\{0,1\}$. This shows that $\{0,1\}^{\mathbb{N}}$ is an uncountable discrete space and thus non-compact. Hence in box topology, even a countably infinite product of $2$-point sets fails to be compact.

A further note: $\{0,1\}^{\mathbb{N}}$ is not even Lindelöf in the box topology. Consider for example the open uncountable cover of singletons, which has no countable (and thus no finite also) subcover.

Consider $\mathbb{R}^\omega$, the countably infinite product of $\mathbb{R}$ with itself. That is,
$$\mathbb{R}^{\omega} = \prod_{n \in \mathbb{Z}_{+}}X_{n}$$
where $X_{n} = \mathbb{R}$ for each $n$. Let us define a function $f: \mathbb{R} \rightarrow \mathbb{R}^{\omega}$ by the equation
$$f(t)=(t,t,t,...)$$
the $n^{\text{th}}$ coordinate function of $f$ is the function $f_{n}(t)=t$. Each of the coordinate functions $f_{n}: \mathbb{R} \rightarrow \mathbb{R}$ is continuous; therefore, the function $f$ is continuous if $\mathbb{R}^\omega$ is given the product topology. But $f$ is not continuous if $\mathbb{R}^\omega$ is given the box topology. Consider, for example, the basis element
$$B = (-1,1) \times \left(-\frac{1}{2},\frac{1}{2}\right) \times \left(-\frac{1}{3},\frac{1}{3}\right) \times \cdot\cdot\cdot$$
for the box topology. We assert that $f^{-1}(B)$ is not open in $\mathbb{R}$. If $f^{-1}(B)$ were open in $\mathbb{R}$, it would contain some interval $(-\delta,\delta)$ about the point $0$. This would mean that $f((-\delta,\delta)) \subset B$, so that applying $\pi_{n}$ to both sides of the inclusion,
$$f_{n}((-\delta,\delta)) = (-\delta,\delta) \subset \left(-\frac{1}{n},\frac{1}{n}\right)$$
for all $n$, which is a contradiction.

Hopefully this is the sort of explicit example you were looking for that compares the product topology and the box topology.

1 - The product topology (PT) is the coarsest topology on a product space that makes every projection continuous. i.e. the PT is the topology generated by the projections. In that sense, PT seems the most natural topology on products (not only because of the name...).

2 - Also, PT tend to have a smaller cardinality then box topologies (BT) (especially if the space is big.). Naturally, we prefer smaller topologies.

3 - If you're in the PT, an open set is basically identified with the finite collection of the open sets that are not the whole space. So an open set in PT is "a finite collection" whence in BT stuff can get pretty wild.

I agree that 1 is very important and that is what we can get immediately from the definition (of subbasis). I think 2 is a good way to describe niceness of PT. I rather prefer to say "product topology as target makes maps easier to be continuous." I.e., it is not natural to me that we just want smaller topologies as it depends on the situations.
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user123454321Oct 24 '12 at 22:33

I think 3 does not say much more information, either. If you can, give me some nice examples as I asked for examples, not general facts.
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user123454321Oct 24 '12 at 22:35